0
$\begingroup$

This question already has an answer here:

if observer 1 sees observer 2 moving toward him at near the speed of light, observer 1 will say that his clock runs faster then observer's 2 clock, but observer 2 will see observer 1 toward him at near the speed of light, and observer 2 will say that observer 1 clock moves slower than his.

when they meet and compare clocks, what will the outcome be?

$\endgroup$

marked as duplicate by John Rennie special-relativity Apr 6 '18 at 18:55

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ This is the famous twin paradox. See physics.stackexchange.com/questions/242043/… $\endgroup$ – bapowell Apr 6 '18 at 13:55
  • 2
    $\begingroup$ In order to make any useful comparison they must meet twice: if they meet only once, as in your scenario, they just can establish that at one particular event (where they meet) their clocks had two particular values, on which they both agree. $\endgroup$ – tfb Apr 6 '18 at 13:59
  • 1
    $\begingroup$ What tfb said. And to meet twice there has to be some acceleration in at least one of the observer's trajectories (assuming simple flattish spacetime) which leads us to the twin paradox. $\endgroup$ – PM 2Ring Apr 6 '18 at 14:18